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Solutions for Session 2, Part H
See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6| H7 | H8
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Problem H1 | |
a. | It is possible. For example:
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b. | It is possible. For example:
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c. | It is possible. For example:
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d. | It is possible. For example:
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e. | It is possible. For example:
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f. | This is impossible, because equal sides correspond to equal angles. This would mean that a triangle would have two consecutive obtuse angles. The sides extending from these two angles could not be connected for the same reason as in Problem A5, part (d).
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g. | It is possible. In fact, all equilateral triangles are also isosceles triangles. |
<< back to Problem H1
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Problem H2 | |
a. | Constructions and answers will vary. But, if the diagonals have different lengths (such as 2 and 3 units), then the figure can never be a square or a rectangle. |
b. | Constructions will vary, but the figure will always be a parallelogram. One possible example is the following:
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c. | Constructions will vary, but the figure will either be a kite or a random quadrilateral. One possible example follows:
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d. | In this case, there is only one possibility -- a rhombus:
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<< back to Problem H2
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Problem H3 | |
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You should find that the result must be a rectangle. Here's one explanation for that fact: The sum of the interior angles in any quadrilateral is 360°. Because the line segments marked "a" are all equal, the angles opposite to them inside the respective triangles are equal. Therefore, the sum of angles 4A + 4B = 360°; i.e., 4(A + B) = 360°; i.e., A + B = 90°. Hence, all of the interior angles are right angles, and the quadrilateral is indeed a rectangle.
<< back to Problem H3
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Problem H4 | |
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The quadrilateral in question is a rectangle as described in the solution to
Problem H3. In addition, two adjacent isosceles right triangles with
hypotenuses a and b respectively are congruent since they have congruent
legs, and the congruent (right) angles between them. So we must have a = b,
and therefore the quadrilateral is a square.
<< back to Problem H4
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Problem H5 | |
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In parts a-c, it is impossible to draw two different triangles. In other words, if we fix two sides and the angle between them, we uniquely determine a triangle.
<< back to Problem H5
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Problem H6 | |
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This type of congruence can be called SAS (side-angle-side) congruence: If two triangles have two sides equal in length, and the angles between those sides are equal in their degree measure, then the two triangles are congruent.
<< back to Problem H6
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Problem H7 | |
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In parts a-c, we can create two or more distinct triangles by keeping the angles fixed and changing the lengths of the sides. For example, we can build a second triangle where each side is twice as long as the original and the angles will remain the same size.
<< back to Problem H7
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Problem H8 | |
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No. Problem H7 shows that two triangles can have the same size angles without being congruent. The triangles appear to have the same overall shape, but they might be larger or smaller than the original. They are not congruent, but they do have a relationship. They are called similar.
<< back to Problem H8
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